Problem: $f(x,y) = y^2\sin(y)$ What is $\dfrac{\partial f}{\partial x}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $2y + \cos(y)$ (Choice B) B $0$ (Choice C) C $2y\cos(y)$ (Choice D) D $2y\sin(y) + y^2\cos(y)$
Solution: We want to find $\dfrac{\partial f}{\partial x}$, which is the partial derivative of $f$ with respect to $x$. When we take a partial derivative with respect to $x$, we treat $y$ as if it were a constant. Here, $f(x, y)$ only has one term. $\begin{aligned} &\dfrac{\partial}{\partial x} \left[ y^2\sin(y) \right] = y^2\sin(y) \dfrac{\partial}{\partial x} \left[ 1 \right] = 0 \end{aligned}$ In conclusion: $\dfrac{\partial f}{\partial x} = 0$